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Mixed domain -- Fourier finite-differences

The pseudo-screen solution to equation 13 derives from a fourth-order expansion of the square-root around $(a_0,b_0)$ and $(a,b)$: k_&& a [1+12( b k_a )^2+ 18( b k_a )^4],
k__0&& a_0[1+12( b_0k_a_0 )^2+ 18( b_0k_a_0 )^4]. If we subtract equations A-16 and A-17, we obtain the following expression for the wavenumber along the extrapolation direction $k_\tau $: k_k__0+ (a-a_0) &+&12[a(b a )^2- a_0(b_0a_0)^2]( k_ )^2
&+&18[a(b a )^4- a_0(b_0a_0)^4]( k_ )^4. We can make the notations _1 &=& a(b a )^2- a_0(b_0a_0)^2,
_2 &=& a(b a )^4- a_0(b_0a_0)^4, therefore equation A-18 can be written as
\begin{displaymath}
k_\tau = {k_\tau }_0+ \omega \left (a-a_0\right )
+ \frac{1}...
...ga \delta_2 \left ( \frac{ k_\gamma }{ \omega } \right )^4\;.
\end{displaymath} (20)

Using the approximation

\begin{displaymath}
\frac{1}{2}\delta_1 u^2 + \frac{1}{8}\delta_2 u^4 \approx
\...
...{1}{2}\delta_1^2 u^2}
{\delta_1-\frac{1}{4}\delta_2 u^2} \;,
\end{displaymath} (21)

we can write
\begin{displaymath}
k_\tau = {k_\tau }_0+ \omega \left (a-a_0\right )
+\omega \f...
...}\delta_2 \left ( \frac{ k_\gamma }{ \omega } \right )^2} \;.
\end{displaymath} (22)

If we make the notations &=& 12_1^2 ,
&=& _1 ,
&=& 14_2 , we obtain the mixed-domain Fourier finite-differences solution to the one-way wave equation in Riemannian coordinates:

\begin{displaymath}
k_\tau \approx {k_\tau }_0+ \omega \left (a-a_0\right )+
\...
...{\mu-\rho \left ( \frac{ k_\gamma }{ \omega } \right )^2} \;.
\end{displaymath} (23)


next up previous [pdf]

Next: Bibliography Up: Sava and Fomel: Riemannian Previous: Mixed domain

2008-12-02